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Q.Find the inverse of matrix [[4,5],[2,3]] by using elementary operations.

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2018Subjective· 4mImportance★★★★★
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Main: A−1=[3/2−5/2−12]A^{-1}=\begin{bmatrix}3/2&-5/2\\-1&2\end{bmatrix}. OR: the determinant equals −2(x3+y3)-2(x^3+y^3).

Main part — inverse by elementary operations. Write A=IAA=IA:

[4523]=[1001]A.\begin{bmatrix}4&5\\2&3\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}A.

R1→R1−R2R_1\to R_1-R_2:

[2223]=[1−101]A.\begin{bmatrix}2&2\\2&3\end{bmatrix}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}A.

R1→12R1R_1\to\tfrac12 R_1:

[1123]=[1/2−1/201]A.\begin{bmatrix}1&1\\2&3\end{bmatrix}=\begin{bmatrix}1/2&-1/2\\0&1\end{bmatrix}A.

R2→R2−2R1R_2\to R_2-2R_1:

[1101]=[1/2−1/2−12]A.\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}1/2&-1/2\\-1&2\end{bmatrix}A.

R1→R1−R2R_1\to R_1-R_2:

[1001]=[3/2−5/2−12]A.\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}3/2&-5/2\\-1&2\end{bmatrix}A.

∴ A−1=[3/2−5/2−12].\therefore\ A^{-1}=\begin{bmatrix}3/2&-5/2\\-1&2\end{bmatrix}.

OR part — evaluate ∣xyx+yyx+yxx+yxy∣\begin{vmatrix}x&y&x+y\\y&x+y&x\\x+y&x&y\end{vmatrix}. …

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